4.4: Inverse Functions

An inverse function is a function that undoes another function: If an input $x$ into the function $f$ produces an output $y$, then putting $y$ into the inverse function $g$ produces the output $x$, and vice versa.

Definition: Inverse Functions

Let $f(x)$ be a 1-1 function then $g(x)$ is an inverse function of $f(x)$ if

$$f(g(x)) = g(f(x)) = x. \nonumber$$

Example $\PageIndex{1}$

For

$$f(x) = 2x - 1 \nonumber$$

$$f^{ -1}(x) = \dfrac{1}{2} x +\dfrac{1}{2} \nonumber$$

since

$$f(f^{ -1}(x) ) = 2 \left( \dfrac{1}{2} x +\dfrac{1}{2} \right) - 1 = x \nonumber$$

and

$$f ^{-1}(f(x)) = \dfrac{1}{2} [2x - 1] + \dfrac{1}{2} = x. \nonumber$$

The Horizontal Line Test and Roll's Theorem

Note that if $f(x)$ is differentiable and the horizontal line test fails then

$$f(a) = f(b) \nonumber$$

and Rolls theorem implies that there is a $c$ such that

$$f '(c) = 0. \nonumber$$

A partial converse is also true:

Theorem $\PageIndex{1}$: Roll's Theorem

If $f$ is differentiable and $f '(x)$ is always non-negative (or always non-positive) then $f(x)$ has an inverse.

Example $\PageIndex{2}$

$$f(x) = x^3 + x - 4 \nonumber$$

has an inverse since

$$f'(x) = 3x^2 + 1 \nonumber$$

which is always positive.

Continuity and Differentiability of the Inverse Function

Theorem $\PageIndex{2}$: Continuity and Differentiability

1. $f$ continuous implies that $f^{ -1}$ is continuous.
2. $f$ increasing implies that $f^{ -1}$ is increasing.
3. $f$ decreasing implies that $f^{ -1}$ is decreasing.
4. $f$ differentiable at $c$ and $f '(c) \neq 0$ implies that $f^{ -1}$ is differentiable at $f (c)$.
5. If $g(x)$ is the inverse of the differentiable $f(x)$ then

$$g'(x) = \dfrac{1}{f '(g(x))}. \lable{Rolls} \nonumber$$

if $f '(g(x)) \neq 0$.

Proof

Since

$$f (g(x)) = x \nonumber$$

we differentiate implicitly:

$$\dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} x.\nonumber$$

Using the chain rule

$$y =f(u), u = g(x)\nonumber$$

$$\begin{align*} \dfrac{dy}{x} &= \dfrac{dy}{dy} \dfrac{du}{dx} \nonumber \\[4pt] &= f '(u) g'(x) = f '(g(x)) g'(x). \end{align*}$$

So that

$$f '(g(x)) g'(x) = 1. \nonumber$$

Dividing, we get Equation $\ref{Rolls}$:

$$g'(x) = \dfrac{1}{f'(g(x))}.\nonumber$$

$\square$

Example $\PageIndex{3}$

For $x > 0$, let

$$f(x) = x^2\nonumber$$

and

$$g(x) = \sqrt{x}\nonumber$$

be its inverse, then

$$g'(x) = \dfrac{1}{2\sqrt{x}}.\nonumber$$

Note that

$$\dfrac{d}{dx} \sqrt{x} = \dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}.\nonumber$$

Exercises

1. Let

$$f(x) = x^3 + x - 4.\nonumber$$

Find

$$\dfrac{d}{dx} f^{-1}(-4). \nonumber$$

2. Let

$$f(x) =\int_2^x \dfrac{1}{1+x^3} dx\nonumber$$

Find

$$\dfrac{d}{dx} f^{-1}(0). \nonumber$$

Outside Links

• http://en.Wikipedia.org/wiki/Rolle%27s_theorem
• Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.